What is the derivative of $e^{i\pi}$?

Question

1. I know that $e^{i\pi}$ =-1

2. the derivative of $e^{i\pi}$ is the derivative of - 1 which is 0.

I guess I'm missing a rule or understood something wrong.

Answer 1

The concept of derivatives is about functions. So, since the derivative of $e^x$ is $e^x$, the derivative of $e^x$ at $i\pi$ is $e^{i\pi}=-1$.

But if you mean the constant function $x\mapsto e^{\pi i}$, its derivative is $0$ everywhere.

Answer 2

The derivative of $e^{ix}$ with respect to $x$ is $ie^{ix}$. You can check this directly by differentiating $\cos(x)+i\sin(x)$.

It turns out that the derivative $e^z$ is $e^z$ also for complex values of $z$, but this may require some knowledge of complex analysis.

Answer 3

$e^{i\pi} = -1$

$e^{i\pi}$ is constant number while $e^{ix} $ is a function. The derivative of $e^{ix}$ at $ x = \pi$ is $ie^{i\pi} = -i.$

But clearly, $e^{i\pi}$ is a constant, so it's derivative is $0.$

This result clearly follows from the definition of derivative. $dy\over dx$ is the rate of change of $y$ with respect to $x$. As the name suggest a constant doesn't depend upon any parameters and hence doesn't change at all. So, the derivative of a constant is $0.$

Answer 4

$$(e^{ax})'=ae^{ax}$$ for any constant $a$, even complex (the prime $'$ denotes differentiation on $x$).

But

$$(e^a)'=0.$$